In the nuclear reaction _3Li^6 + (?) to _2He^ | vedclass

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(B) To find the missing particle,we balance the mass numbers and atomic numbers on both sides of the equation.Let the missing particle be $_Z^A X$.The

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Neutron

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(B) To find the missing particle,we balance the mass numbers and atomic numbers on both sides of the equation.Let the missing particle be $_Z^A X$.The equation is $_3Li^6 + _Z^A X 	o _2He^4 + _1H^3$.Sum of mass numbers: $6 + A = 4 + 3 \implies 6 + A = 7 \implies A = 1$.Sum of atomic numbers: $3 + Z = 2 + 1 \implies 3 + Z = 3 \implies Z = 0$.$A$ particle with mass number $1$ and atomic number $0$ is a neutron,denoted as $_0n^1$.Therefore,the missing particle is a neutron.

In the nuclear reaction $_3Li^6 + (?) \to _2He^4 + _1H^3$,the missing particle is:

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